## Precalculus (6th Edition) Blitzer

The limit $\underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)$ is equal to $-1$.
Consider the provided limit, $\underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)$ Simplify it as follows \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x\times 1-x\times \frac{1}{x} \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,\left( x-1 \right) \end{align} Use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x-1 \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,x-\underset{x\to 0}{\mathop{\lim }}\,1 \end{align} Using limit property, $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c\text{ is constant}$ \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x-1 \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,x-\underset{x\to 0}{\mathop{\lim }}\,1 \\ & =0-1 \\ & =-1 \end{align} Thus, $\underset{x\to 0}{\mathop{\lim }}\,x\left( 1-\frac{1}{x} \right)=-1$